Let $R$ be a ring. Let $I\lhd R$ and fix $n\in I$ if $n$ is the unit of $R$. Show $R=I$

Question

Let $R$ be a ring. Let $I\lhd R$ (that is $I$ is an ideal of the ring) and fix $n\in I$ if $n$ is the unit of $R$. Show $R=I$.

Here is my attempt at an answer:

We aim to show $I \subseteq R$ and $R \subseteq I$ to do this pick an element $m \in I$ and then it follows $m \in R$ by definition of ideals.

Now let $k \in R$ then $kn=nk=k$ and $kn,nk \in I$ since $k \in R$ and $n \in I$ and by definition of ideals.

Hence $I \subseteq R$ and $R \subseteq I$ that is $I=R$ $~~~~~~~\blacktriangle$

Is this a good answer?

Answer 1

No, this is incorrect. To say that $n$ is a unit means that there is another element $m$ in the ring for which $nm = mn = 1_R$, where $1_R$ is the identity of the ring. For example, every non-zero element in $\mathbb{Q}$ is a unit in $\mathbb{Q}$, but the equation $nk = k$ can only hold for exactly one value of $n$.

So what you can conclude is that $1_R \in I$. And once you have that, you can apply some of the steps you wrote above to show that $R \subseteq I$.